The Unapologetic Mathematician

Mathematics for the interested outsider

The Monotone Convergence Theorem

We want to prove a strengthening of the dominated convergence theorem. If \{f_n\} is an a.e. increasing sequence of extended real-valued, non-negative, measurable functions, and if f_n converges to f pointwise a.e., then

\displaystyle\lim\limits_{n\to\infty}\int f_n\,d\mu=\int f\,d\mu

If f is integrable, then f dominates the sequence \{f_n\}, and so the dominated convergence theorem itself gives us the result we assert. What we have to show is that if \int f\,d\mu=\infty, then the limit diverges to infinity. Or, contrapositively, if the limit doesn’t diverge then f must be integrable.

But this is the limit of a sequence of real numbers, and so if it converges then it’s Cauchy. That is, we can conclude that

\displaystyle\lim\limits_{m,n\to\infty}\left\lvert\int f_m\,d\mu-\int f_n\,d\mu\right\rvert=0

Our assumption that f_n is a.e. increasing tells us that for any fixed m and n, the difference f_m-f_n is either a.e. non-negative or a.e. non-positive. That is,

\displaystyle\left\lvert\int f_m\,d\mu-\int f_n\,d\mu\right\rvert=\int\lvert f_m-f_n\rvert\,d\mu

And thus the sequence \{f_n\} is mean Cauchy, and thus mean convergent to some integrable function g, which must be equal to f almost everywhere.

One nice use of this is when talking about series of functions. If \{f_n\} is a sequence of integrable functions so that

\displaystyle\sum\limits_{n=1}^\infty\int\lvert f_n\rvert\,d\mu<\infty

then I say that the series

\displaystyle\sum\limits_{n=1}^\infty f_n(x)

converges a.e. to an integrable function f, and further that

\displaystyle\int f\,d\mu=\int\sum\limits_{n=1}^\infty f_n\,d\mu=\sum\limits_{n=1}^\infty\int f_n\,d\mu

To see this, we let S_n(x) be the partial sum

\displaystyle S_n(x)=\sum\limits_{i=1}^n\lvert f_i(x)\rvert

which gives us a pointwise increasing sequence of non-negative measurable functions. The monotone convergence theorem tells us that these partial sums converge pointwise to some S(x) and that

\displaystyle\int S\,d\mu=\lim\limits_{n\to\infty}\int S_n\,d\mu=\lim\limits_{n\to\infty}\int\sum\limits_{i=1}^n\lvert f_i\rvert\,d\mu=\lim\limits_{n\to\infty}\sum\limits_{i=1}^n\int\lvert f_i\rvert\,d\mu=\sum\limits_{i=1}^\infty\int\lvert f_i\rvert\,d\mu

But this is exactly the sum we assumed to converge before. Thus the function S is integrable and the series of the f_n is absolutely convergent. That is, since S must be a.e. finite, the series

\displaystyle\sum\limits_{n=1}^\infty f_n(x)

is absolutely convergent for almost all x, and so it must be convergent pointwise almost everywhere. Since S dominates the partial sums

\displaystyle S(x)=\sum\limits_{i=1}^\infty\lvert f_i(x)\rvert\geq\sum\limits_{i=1}^n\lvert f_i(x)\rvert\geq\left\lvert\sum\limits_{i=1}^nf_i(x)\right\rvert

the bounded convergence theorem tells us that limits commute with integrations here, and thus that

\displaystyle\begin{aligned}\int f\,d\mu&=\int\sum\limits_{i=1}^\infty f_i\,d\mu\\&=\int\lim\limits_{n\to\infty}\sum\limits_{i=1}^nf_i\,d\mu\\&=\lim\limits_{n\to\infty}\int\sum\limits_{i=1}^nf_i\,d\mu\\&=\lim\limits_{n\to\infty}\sum\limits_{i=1}^n\int f_i\,d\mu\\&=\sum\limits_{i=1}^\infty\int f_i\,d\mu\end{aligned}

June 15, 2010 Posted by | Analysis, Measure Theory | 10 Comments

   

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